1 Section R.4 Review of Factoring Objective #1: Factoring Out the Greatest Common Factor The Greatest Common Factor (GCF) is the largest factor that can divide into the terms of an expression evenly with no remainder. To find the GCF of variables, list the variables that appear in all the terms and then choose the lowest power of each respective variable as the power for the GCF. Ex. 1a 3p 3 15p 2 Ex. 1b 42x 3 y 2 z + 14xy 2 z 2 35x 4 y 4 Ex. 1c 12p 3 t + 2p 2 t 3 + 6pt 2 2pt Ex. 1d 9x(2x + 3) 6(2x + 3) a) Since the GCF = 3p 2, then 3p 3 15p 2 = 3p 2 (p) 3p 2 (5) = 3p 2 (p 5) b) Since the GCF = 7xy 2, then 42x 3 y 2 z + 14xy 2 z 2 35x 4 y 4 = 7xy 2 (6x 2 z) + 7xy 2 (2z 2 ) 7xy 2 (5x 3 y 2 ) = 7xy 2 (6x 2 z + 2z 2 5x 3 y 2 ) c) Since the GCF = 2pt, then 12p 3 t + 2p 2 t 3 + 6pt 2 2pt = 2pt(6p 2 ) 2pt( pt 2 ) 2pt(3t) 2pt(1) = 2pt(6p 2 pt 2 3t + 1) d) Since the GCF = 3(2x + 3), then 9x(2x + 3) 6(2x + 3) = 3(2x + 3) 3x 3(2x + 3) 2 = 3(2x + 3)(3x 2) Objective #2: Factoring by Grouping Whenever a factoring problem has four or more terms, we put the terms into groups and factor out the G.C.F. of each group. We then see if the groupings have any pieces in common. If so, we then factor out the common pieces. Ex. 2a my + 6y 7m 42 Ex. 2b w 2 + 5w + 20z + 4wz my + 6y 7m 42 w 2 + 5w + 20z + 4wz = my + 6y + 7m 42 = (w 2 + 5w) + (20z + 4wz) = (my + 6y ) + ( 7m 42) = w(w + 5) + 4z(5 + w) = y(m + 6) 7(m + 6) = w(w + 5) + 4z(w + 5) = (m + 6)(y 7) = (w + 5)(w + 4z)
2 Ex. 2c 10uv 3 5v 3 30uv 2 + 15v 2 Factor out GCF first. 10uv 3 5v 3 30uv 2 + 15v 2 = 5v 2 [2uv v 6u + 3] = 5v 2 [(2uv v) + ( 6u + 3)] = 5v 2 [v(2u 1) 3(2u 1)] = 5v 2 (2u 1)(v 3) Objective #3: Factoring Trinomials in the form x 2 + bx + c To factor trinomials where the coefficient of the square term is 1, we will look at the factors of c to see which pair of factors add up to b. We will then use that result to rewrite the problem as a product of two binomials. Factoring x 2 + bx + c 1) List all pairs of factors of c. 2) Determine which pair of factors e 1 and e 2 found in part 1 you can add to get b. If you get the opposite of b for the sum, change the signs of e 1 and e 2. If a pair factors cannot be found, then the trinomial is said to be "prime." 3) Rewrite the trinomial as (x + e 1 )(x + e 2 ). If the original trinomial was in the form x 2 + bxy + cy 2, then the result will be (x + e 1 y)(x + e 2 y). 4) Verify that the product gives you the original trinomial. Ex. 3a 25x 3 275x 2 y + 600xy 2 The G.C.F. = 25x, so 25x 3 275x 2 + 600x = 25x(x 2 11xy + 24y 2 ). We want to find a pair of factors of 24 that add up to 11. Pairs of factors of 24 b = Sum of factors of c = 11 1 24 1 + 24 = 25 No 2 12 2 + 12 = 14 No 3 8 3 + 8 = 11 Almost, change the signs 4 6 4 + 6 = 10 No So, 3 and 8 are what we want to use. Since 24 had a y 2 next to it, our answer will be 25x(x 3y)(x 8y).
Ex. 3b 4a 3 b 2 + 40a 2 b 3 + 224ab 4 G.C.F. = 4ab 2, so 4a 3 b 2 + 40a 2 b 3 + 224ab 4 = 4ab 2 (a 2 10ab 56b 2 ) Pairs of factors of 56 b = Sum of factors of c = 10 1 56 1 + 56 = 55 No 2 28 2 + 28 = 26 No 4 14 4 + 14 = 10 Almost, change the signs 7 8 7 + 8 = 1 No So, 4 and 14 are what we want to use. Since 56 had a b 2 next to it, our answer will be 4ab 2 (a 14b)(a + 4b). 3 Objective #4: Factoring Trinomials Using AC Method. To factor trinomials, we will use a method called the AC method. With this method, you uncombine the middle term and then factor the problem by grouping. Here is the procedure: AC Method 1) Multiply the coefficient of the first and last terms (a c). 2) List all pairs of factors of result from step #1 until you find a pair whose sum is the coefficient of the middle term. 3) Rewrite the middle term as the sum of two terms using the factors found in step #2 as the coefficients. 4) Factor by grouping. Ex. 4a 3x 2 5x 12 The coefficients of the first and last terms are 3 and 12. Their product is 36. factors of 36 sum = 5 1 36 1 + 36 = 35 No 2 18 2 + 18 = 16 No 3 12 3 + 12 = 9 No 4 9 4 + 9 = 5 Yes, change the signs 6 6 Rewrite 5x as 4x 9x 3x 2 5x 12 = 3x 2 + 4x 9x 12 = (3x 2 + 4x) + ( 9x 12) = x(3x + 4) 3(3x + 4) = (3x + 4)(x 3)
Ex. 4b 135x 2 + 9xy + 54y 2 The G.C.F. = 9, so 135x 2 + 9xy + 54y 2 = 9(15x 2 xy 6y 2 ). The coefficients of the first and last terms are 15 and 6. Their product is 90. factors of 90 sum = 1 1 90 1 + 90 = 89 No 2 45 2 + 45 = 43 No 3 30 3 + 30 = 27 No 5 18 5 + 18 = 13 No 6 15 6 + 15 = 9 No 9 10 9 + 10 = 1 Almost, change the signs Ex. 4c 4a 3 b 25a 2 b 2 30ab 3 G.C.F. = ab, so 4a 3 b 25a 2 b 2 30ab 3 = ab(4a 2 + 25ab + 30b 2 ) The coefficients of the first and last terms are 4 and 30. Their product is 120. factors of 120 Sum = 25 1 120 1 + 120 = 121 No 2 60 2 + 60 = 62 No 3 40 3 + 40 = 43 No 4 30 4 + 30 = 34 No 5 24 5 + 24 = 29 No 6 20 6 + 20 = 26 No 8 15 8 + 15 = 23 No 10 12 10 + 12 = 22 No Since there are no pairs that yield a sum of 25, the trinomial is prime. Thus, our answer is ab(4a 2 + 25ab + 30b 2 ). 4 Rewrite xy = 9xy 10xy, 9(15x 2 xy 6y 2 ) = 9(15x 2 + 9xy 10xy 6y 2 ) = 9([15x 2 + 9xy] + [ 10xy 6y 2 ]) = 9(3x[5x + 3y] 2y[5x + 3y]) = 9(5x + 3y)(3x 2y) Objective #5: Factoring the Difference of Squares To factor a binomial that is a difference of squares, we can use the following special product: F 2 L 2 = (F L)(F + L) In order for a binomial to fit the pattern, the first term has to be a perfect square, the last term has to be a perfect square, and the operation has to be subtraction.
5 Ex. 5a 9x 2 16y 2 Ex. 5b 49y 3 36y 9x 2 16y 2 49y 3 36y = (3x) 2 (4y) 2 = y(49y 2 36) = y([7y] 2 [6] 2 ) F = 3x, L = 4y and operation is. F = 7y, L = 6 and operation is. 9x 2 16y 2 = (3x 4y)(3x + 4y) y(49y 2 36) = y(7y 6)(7y + 6) Ex. 5c 81x 2 + 25z 2 Ex. 5d 16x 4 81 The operation is + so this cannot 16x 4 81 = (4x 2 ) 2 (9) 2 be factored. 81x 2 + 25z 2 is prime*. F = 4x 2, L = 9 and operation is. 16x 4 81 = (4x 2 9)(4x 2 + 9) * - Note that 81x 2 + 25z 2 does = ([2x] 2 [3] 2 )(4x 2 + 9) factor in the complex numbers. If F = 2x, L = 3 and operation is. we let i = 1, so i 2 = 1, then = (2x 3)(2x + 3)(4x 2 + 9) 81x 2 + 25z 2 = (9x 5i z)(9x + 5i z). Objective #6: Factoring a Perfect Square Trinomial To factor a trinomial that is a perfect, we can use the following special products: 1) F 2 + 2FL + L 2 = (F + L) 2 2) F 2 2FL + L 2 = (F L) 2 For a trinomial to fit the pattern, the first term has to be a perfect square, the last term has to be a perfect square, and the middle term has to match. Ex. 6a 49x 2 + 56xy + 16y 2 Ex. 6b 20a 3 60a 2 + 45a F = 7x, L = 4y and 2FL = 56xy. 20a 3 60a 2 + 9a = 5a(4a 2 12a + 9) = (7x) 2 + 2(7x)(4y) + (4y) 2 F = 2a, L = 3 and 2FL = 12a. 49x 2 + 56xy + 16y 2 = (7x + 4y) 2 = 5a([2a] 2 2[2a][3] + [3] 2 ) 5a(4a 2 12a + 9) = 5a(2a 3) 2 Ex. 6c 100x 2 + 360xz + 324z 2 Ex. 6d 4w 2 25w + 36 = 4(25x 2 + 90xz + 81z 2 ) = 4([5x] 2 + 2[5x][9z] + [9z] 2 ) F = 2w, L = 6, but 2FL = 24w. F = 5x, L = 9z and 2FL = 90xz So, the pattern does not work. We 4(25x 2 + 90xz + 81z 2 ) = 4(5x + 9z) 2 will have to factor using the AC Method to get (4w 9)(w 4).
Objective #7: Factoring the Sum or Difference of Cubes To factor a binomial that is a sum or difference of cubes, we can use the following special products: F 3 + L 3 = (F + L)(F 2 FL + L 2 ) * - if the degree of F 2 FL + L 2 and F 3 L 3 = (F L)(F 2 + FL + L 2 ) F 2 + FL + L 2 is 2, then they are prime in the real numbers. In order for a binomial to fit the pattern, the first term has to be a perfect cube and the last term has to be a perfect cube. Ex. 7a 27x 3 64y 3 27x 3 64y 3 = (3x) 3 (4y) 3 F = 3x, L = 4y and operation is. = (3x 4y)([3x] 2 + [3x][4y] + [4y] 2 )= (3x 4y)(9x 2 + 12xy + 16y 2 ) Ex. 7b 125a 3 + 8b 3 125a 3 + 8b 3 = (5a) 3 + (2b) 3 F = 5a, L = 2b and operation is +. = (5a + 2b)([5a] 2 [5a][2b] + [2b] 2 ) = (5a + 2b)(25a 2 10ab + 4b 2 ) Objective #8: Factoring Using Substitution Sometimes is difficult to factor a problem directly. What we can do is to notice a pattern, make a substitution into the problem, which gives us an easier problem to factoring. After factoring the easier problem, we substitute back in the original expression. Ex. 8 4x 4 63x 2 16 Let u = x 2. Our problem becomes: 4x 4 63x 2 16 = 4u 2 63u 16 The coefficients of the first and last terms are 4 and 16. Their product is 64. Using the AC Method, we get: factors of 64 sum = 63 1 64 1 + 64 = 63 Almost, change the signs Now, replace u = x 2 and simplify: (4u + 1)(u 16) = (4x 2 + 1)(x 2 16) But, x 2 16 = (x 4)(x + 4), so, (4x 2 + 1)(x 2 16) = (4x 2 + 1)(x 4)(x + 4) Rewrite 63u = u 64u, 4u 2 63u 16 = 4u 2 + u 64u 16 = u[4u + 1] 16[4u 1] = (4u + 1)(u 16) 6